Question

Number of zero's in the end of the product of $5^6\times 6^7\times 7^8\times \cdots \times {30}^{31}$ is

• A. $111$
• B. $147$
• C. $142$
• D. $137$

Answer 1

The correct option is D $137$

One pair of $2$ and $5$ makes $10$
(i.e., $2\times 5=10$) which gives one zero at the end.
Now
$5^6,{10}^{11},{15}^{16},{20}^{21},{25}^{26},{30}^{31}$ contain $6,11,16,21,2\times 26,31$ fives.
Therefore total number of fives
$=6+11+16+21+2\times 26+31=137$
Also, number of $2^{\prime }s$ is greater than number of $5^{\prime }s$.
Thus, number of zero at the end is $137$.